HEEELP ME !!!!!!! Determine the equation of each graph in the form y = acos(bx). https://www.virtualhighschool.com/content/OSSD-v2/Department/Math/courses/MCR3U/ver-D/Transforming_Trig/images/45.png?_&d2lSessionVal=zz5owIE4LYpM64Qh3q7GmzDHv and https://www.virtualhighschool.com/content/OSSD-v2/Department/Math/courses/MCR3U/ver-D/Transforming_Trig/images/46.png?_&d2lSessionVal=zz5owIE4LYpM64Qh3q7GmzDHv B) Determine the equation of each graph above in the form y = a sin(bx)
PLEASE HELP!!
in the first one it should be more or less clear that \(a=4\) because that is the highest point on the graph
what is the equation then?
\(b\) is found by setting \[\frac{360}{b}=1440\] and solving for \(b\) since the period of that function is \(1440\)
i.e. \(b=\frac{360}{1440}=\frac{1}{4}\)
wont b = 4?
@tkhunny can you help me?
b=4??
no it will not be 4, it will be \(\frac{1}{4}\) for sure
ok
so the equation will be y= 4cos(1/4x) ?
\[\frac{360}{b}=1440\iff 1440x=360\iff x=\frac{360}{1440}=\frac{1}{4}\]
yes or \[4\cos(\frac{x}{4})\]
can you do the second one??
probably let me look
this one looks like sine, but instead of starting at zero and going up, it starts at zero and goes down so \(a\) will be negative
i can't really see how far down it goes what is the lowest point on the graph?
-3
ok so \(a=-3\)
you do not see a full period here but you can figure it out from the picture, because half of the period is 90 degrees therefore the full period is 180 degrees
ok
to find \(b\) as before, solve \[\frac{360}{b}=180\] for \(b\)
the equation is -3cos(2x) ???
not cosine, sine
if it is right ...... can you do part b?
ok
i thought that was B
no you have to write the cos equation for the two graphs and then part b is writing the sine for the same two graphs
@tkhunny help!!
Join our real-time social learning platform and learn together with your friends!